For the circuit shown in the figure, find the magnitude and the direction of the current through

For the circuit shown in the figure, find the magnitude and...

Key Concepts

Kirchhoff's Voltage Law (KVL)

This law states that the sum of the voltage drops and rises around any closed loop in a circuit is zero. It is fundamental in setting up equations for loop (mesh) analysis, allowing the determination of unknown currents when resistors and voltage sources are present in the circuit.

Kirchhoff's Current Law (KCL)

KCL asserts that the total current entering a junction (or node) equals the total current leaving the junction. This principle is useful in establishing relationships between branch currents at nodes, which is crucial for solving complex circuits with multiple interconnected branches.

Ohm's Law

Ohm's Law defines the linear relationship between voltage, current, and resistance in an electrical circuit by the equation V = IR. It is essential for calculating voltage drops across resistive elements and determining the magnitude of currents in various branches of the circuit.

Resistive Network Analysis

This concept involves understanding and simplifying resistor combinations in series and parallel configurations. By calculating equivalent resistances, one can reduce complex circuits into simpler forms, making the analysis of current distributions and voltage divisions more manageable.

Mesh Analysis

Mesh analysis is a systematic technique used to solve for unknown currents in planar circuits. By defining independent loop currents and applying KVL to each loop, one creates a set of simultaneous equations that can be solved to determine the current through each resistor.

Power Calculation in Circuits

This concept refers to the determination of the power associated with circuit elements, such as the power delivered by batteries or consumed by resistors. Power is typically calculated using the formula P = IV, where careful attention must be given to the sign convention, which indicates whether the component is supplying or consuming power.

Transcript

00:01 So here we have network of the resistors and we will need to find out the current flowing through each resistors.

00:08 So let's say the current flowing is i1 in this branch and i2 in this branch.

00:16 So applying kirchop's current law, we will get the current flowing through this 5 om as i1 minus i2.

00:24 So similarly here the current flowing will be i2 and we take the current flowing through the 6 oom as i 3 so the current flowing through the 6 om will be i 2 minus i 3 and here these two currents will combine and the total current flowing through the 6 volt will be i 1 minus i 2 plus i 3 and here as usual i 1 current goes out of the battery so i 1 current will come out here so we will apply kirchof's voltage law across three loops which because we have three different unknown variables.

01:09 So let us say first we apply in this loop.

01:12 So here we start from this point.

01:16 We take it as zero potential.

01:18 So potential at this point will be 12.

01:20 And potential at this point will be decreased by this 5 -oom resistor.

01:27 And then we are moving from positive to negative.

01:30 So there will be decrement of 6 volt.

01:32 And then this 10 and 3 are in series.

01:35 There will be decrease in potential due to this two resistors.

01:38 So 13 i1 and this will be 0 so this will be the first equation so let us simplify it even more so taking 5 i1 and 13 i1s we'll get minus 18 i1 plus 5 i2 plus 6 equals to 0 so this is first equation now we consider let's say this loop so we start at this point with 0 potential and we are moving opposite to the direction of current so potential here will be 5 i1 minus i 2 and then potential here will be minus 8 i2 and here the potential will be minus 4 i2 and similarly now the potential will be minus 6 i 3 and it will be 0 so we'll simplify it even more so we get 5 i 1 minus 5 i 2 minus 8 i 2 that is minus 13 i 2 and then minus 4 i 2 so that we'll make it minus 17 i 2 and minus 6 i 3 as usual so this is equation 2 let's put them in the box so that we are sure that these are the equations so similarly we'll take the third loop let's say i take it this one so we start at this point 0 plus 6 so the third equation from this loop we use kitch jobs voltage log in so 6 minus so here we are going oppose the it to the direction of current so it will be plus 6 i3 minus 6 times i 2 minus i 3 and this will be 0 so simplifying it 6 i 3 plus 6 i 3 here so we get 12 i 3 minus 6 i 2 plus 6 equals to 0 and this is equation 3 so we solve these 3 equations so we'll take 2 equation at a time and get the value of i 1 and i 2 and then substitute in the 3 to get in i 3 so we'll so from the first equation we can write i1 as 5 i2 plus 6 by 18.

04:10 So second equation it is also in terms of i1 and i2 and i3.

04:15 So we again take the value of i3 now and convert it in the form of i2.

04:21 So we get 5.

04:23 Instead of i1 we will write it as 5 i2 plus 6 by 18 minus 17 i2 minus 6 i3.

04:33 Equals to zero so converting the second equation we'll get 25 i2 plus 30 by 18 minus 17 i2 minus 6 i3 is equal to zero so here we need to calculate them a bit because it's involving quite bigger numbers so here we get 25 i2 plus 30 minus 18 times 17 that is 306 i2 minus 18 times 6 that is 108 i 3 equals to 0 so from here we find out i 3 so we combine 25 i 2 and 306 i 2 so that is minus 281 i 2 plus 30 minus 108 i 3 so from here we get i 3 as 30 minus 20 8 i 3 so from here we get i 3 as 30 minus 2 8 181 i2 by one not 8 so we take this value of i1 and i3 in terms of i2 and replace in this third equation so let's replace that we get 12 times i 3 so you'll write 12 times the value of i 3 from here that is 30 minus 281 i2 over 108 and then minus 6 i2 we will keep it as it is and then we have plus 6 is equal to so from here we can figure out i2...

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